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The reason that it's so hard to understand what physicists mean when they talk about "gauge freedom" is that there are at least four inequivalent definitions that I've seen used:
Definition 1: A mathematical theory has a gauge freedom if some of the mathematical degrees of freedom are "redundant" in the sense that two different mathematical expressions ...
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Gauge symmetries, as the note says, are redundancies in our description of nature. For example, a photon has two physical degrees of freedom (the two polarizations). However, we choose to describe a photon using a 1-form field $A_\mu$ which has 4 degrees of freedom. The two extra degrees of freedom here are related to gauge symmetries. From here on, there ...
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You're making some category errors in the question. Energy can't be converted into mass, mass is a form that energy can take. In other words, when energy is "converted" into mass it never stops being energy. It's kind of like if I have a mass on a spring hanging vertically in a gravitational field, and I make it start bouncing. The energy moves back and ...
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The first answer to such a question must always be: A gauge symmetry has no "physical" meaning, it is an artifact of our choice for the coordinates/fields with which we describe the system (cf. Gauge symmetry is not a symmetry?, What is the importance of vector potential not being unique?, "Quantization of gauge systems" by Henneaux and Teitelboim). Any ...
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I only understood this after taking a class in general relativity (GR), differential geometry and quantum field theory (QFT). The essence is just a change of coordinates systems that needs to be reflected in the derivative. I'll explain what I mean.
You have a theory that is invariant under some symmetry group. So in quantum electrodynamics you have a ...
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A theory is typically described by a Lagrangian, and varying this gives us the equations of motion of the system. The symmetries you describe are symmetries of the Lagrangian i.e. they are transformations that leave the Lagrangian unchanged.
It would be nice to think that the Lagrangians that describe our leading theories of physics were derived in some ...
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Since you mentioned coming from a mathematics background, you might find it nice to take an answer in terms of equivalence classes.
A gauge theory is physical theory where the observable quantities, as in, things you could measure with an experiment given perfect measuring equipment, are equivalence classes in a vector space.
Electromagnitism is the most ...
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We do not start from the assumption that the Lagrangian "should" be invariant under gauge transformations. This assumption is often made because global symmetries are seen as more natural than local symmetries and so writers try to motivate gauge theory by "making the global symmetry local", but this is actually nonsense. Why would we want a local symmetry ...
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Bundles and compactified spacetime
A gauge theory cannot be looked at purely locally, it has inherently global features one cannot see locally. The proper mathematical formalization of a Yang-Mills gauge theory is that the gauge field $A$ is a connection on a principal bundle $P\to M$ over spacetime $M$. However, in practice, it turns out that physicists don'...
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Let us first refer to symmetry generically. When we say a theory is symmetric under $G$ ($G$ some group) we mean that the elements of $G$ transform the states, and the operators of a theory, (in the context of the Standard Model (SM)), in such a way that the Lagrangian won't change in form. One could then speak about space-time symmetries, such as Lorentz ...
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The guiding principle is: "Anomalous symmetries cannot be gauged". The phenomenon of anomalies is not confined to quantum field theories. Anomalies exist also in classical field theories (I tried to emphasize this point in my answer on this question).
(As already mentioned in the question), in the classical level, a symmetry is anomalous when the Lie ...
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A free "$\mathrm{U}(1)$" gauge theory can never tell whether the gauge group is $\mathrm{U}(1)$ or $\mathbb{R}$ because the only field in the theory, the gauge potential $A$, transforms as
$$ A\mapsto A + \partial_\mu \chi,$$
where $\chi$ is just a real-valued function, and the real numbers are the Lie algebra of both $\mathrm{U}(1)$ and $\mathbb{R}$. This ...
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I) Vanishing field-strength $F=0$ does not imply that the gauge potential $A$ is pure gauge. It only holds locally. There could be global obstructions. In fact, topological obstructions could happen even if the gauge group $G$ is Abelian.
II) Let us sketched the proof of the local statement in a sufficiently small neighborhood $\Omega\subseteq M$ of a point ...
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The fact that the theory is not gauge invariant implies that all degrees of freedom of $A_\mu$ must have physical meaning: This is not the theory of photons where only transverse degrees of freedom make sense.
This way you must tackle some non-trivial issue like the negative norm associated with temporal modes. This could be avoided by adding a mass to $A_\...
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Here is a simple example, one of the first you should try to understand. The theory has a free $U(1)$ scalar field $\phi$ in $d+1$ spacetime dimensions, discussed in the modern notation of differential forms. The Lagrangian density
$$L_0 = d\phi \wedge\star d\phi.$$
This has a manifest global symmetry $\phi \mapsto \phi + \theta$. If we perform a local ...
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Let there be given a 4-dimensional real manifold$^1$ $M$. As OP says, the set ${\rm Diff}(M)$ is the group of globally defined $C^{\infty}$-diffeomorphisms $f:M\to M$. The set ${\rm Diff}(M)$ is an infinite-dimensional Lie group. (To actually explain mathematically what the previous sentence means, one would have to define what an infinite-dimensional ...
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OP has a point. The field $A^\mu$ is a connection, and therefore it lives in the algebra of the gauge group, not in the group itself. In this case, $\mathfrak u(1)=\mathbb R$. At first sight, this is all we may conclude from $A\to A+\mathrm d\Lambda$. The group $\mathrm U(1)$ is, apparently, not here yet.
The correct statement is that the theory described ...
answered Jan 29 '18 at 15:56
AccidentalFourierTransform
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I) In general, it is true that if we plug a local Lagrangian
$$\tag{1} L\quad \longrightarrow \quad \tilde{L}~=~L+\frac{df}{dt}$$
modified with a total derivative term into the Euler-Lagrange expression
$$\tag{2} \sum_{n} \left(-\frac{d}{dt}\right)^n \frac{\partial \tilde{L}}{\partial q^{(n)}}~=~\sum_{n} \left(-\frac{d}{dt}\right)^n \frac{\partial L}{\...
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Multiplying by $e^{i\theta}$ is a rotation of $\theta$ in the complex plane. Physically it changes the phase of a plane wave by an angle $\theta$. This is a global symmetry because we arbitrarily choose a reference point for measuring the phase of plane waves. If we change the phase of all plane waves by an equal amount then this is equivalent to just moving ...
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Gauge invariance is simply a redundancy in the description of a physical system. I.e. we can choose from an infinite number of vector potentials in E&M.
For example, an infinite number of vector potentials can describe electromagnetism by the transformation below
$$A(x) \to A_\mu(x) + \partial_\mu \alpha(x)$$
Choosing a specific gauge (gauge fixing) ...
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I think your remark 1 is right on, and so I disagree with remark 2.
I do not believe that you should consider the gauge formalism as fundamental or even unavoidable. Gauge degrees of freedom arise when you add in fake degrees of freedom in order to make the formulation of the theory simpler or more manifestly symmetric. For instance, we want the quantum ...
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Comment to the question (v1): It seems OP is conflating, on one hand, a gauge transformation
$$ \tilde{A}_{\mu} ~=~ A_{\mu} +d_{\mu}\Lambda $$
with, on the other hand, a gauge-fixing condition, i.e. choosing a gauge, such e.g., Lorenz gauge, Coulomb gauge, axial gauge, temporal gauge, etc.
A gauge transformation can e.g. go between two gauge-fixing ...
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These calculations very often depend only on the difference between two values, not the concrete values themselves. You are therefore free to choose a zero to your liking. Is this an example of gauge invariance in the same sense as the graduate examples above?
Yes indeed it is, in the most general definition of gauge invariance, it's what physicists call a ...
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You seem to be mixing up a few different things. The Standard Model does not say the up and down quark must have the same mass.
The up and down quark form a doublet of the flavor symmetry $SU(2)_F$. This is an approximate symmetry that is broken explicitly by the up and down quark mass difference.
The left-handed up and down quarks form a doublet $Q_L$ of ...
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Here's the most elementary example of a gauge symmetry I can think of.
Suppose you want to discuss some ants walking around on a Möbius band. To describe the positions of the ants, it's convenient to imagine cutting the band along its width, so it becomes a rectangle. Then you can tell me where an ant is by telling me three things:
Her latitude—her ...
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If you don't impose power-counting renormalizability, there are a host of other possibilities, since higher order derivatives or higher order interactions can be introduced. For example, terms $(Tr(F^2)^m)^n$ and are gauge invariant but for $m>1$ or $n>1$ not renormalizable.
If you impose power-counting renormalizability, uniqueness is fairly ...
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The statement
because the gauge transformation is not supposed to change anything, it means that every expectation can be calculated equivalently using $ \psi' $ or $ \psi $
isn't particularly correct. All physically measurable expectation values can be calculated correctly in any arbitrary gauge, but the operator representation of some operators can ...
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The fields are not affected by this gauge transformations, and only those quantities have physical meaning that are invariant under the transformations. So essentially the physics is the same in these two gauges. But if you want some of the implications of choosing one of these gauges, so you can choose the appropriate one, the following may be of help:
...
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I) The topic of gauging global symmetries is a quite large subject, which is difficult to fit in a Phys.SE answer. Let us for simplicity only consider a single (and thus necessarily Abelian) continuous infinitesimal transformation$^1$
$$ \tag{1} \delta \phi^{\alpha}(x)~=~\varepsilon(x) Y^{\alpha}(\phi(x),x), $$
where $\varepsilon$ is an infinitesimal real ...
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The expression of the Chern-Simons functional as an integral over a 3-apace is just a shorthand notation. The integration in the Chern-Simons functional differs from the integration of differential forms.
The integration can formulated by means of the theory of Deligne-Beilinson cohomology. (Please, see the following very clear review by Frank Thuillier, ...
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